# Abstract Nonsense

## Munkres Chapter 2 Section 19 (Part II)

9.

Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ of non-empty sets, with $\mathcal{A}\ne\varnothing$, the product $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$

Proof: This is pretty immediate when one writes down the actual definition of the product, namely:

$\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=\left\{\bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}:\bold{x}(\alpha)\in U_\alpha,\text{ for every }\alpha\in\mathcal{A}\right\}$

So, if one assumes the AOC then one must assume the existence of a choice function

$\displaystyle c:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }c\left(U_\alpha\right)\in U_\alpha\text{ for all }\alpha\in\mathcal{A}$

So, then if we consider $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}\Omega$ as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$$i:\mathcal{A}\to\Omega$ where clearly since we have already indexed out set we have that $i:\alpha\mapsto U_\alpha$. So, consider $c\circ i:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ This is clearly a well-defined mapping and $\left(c\circ i\right)(\alpha)=c\left(U_\alpha\right)\in U_\alpha$ and thus $\displaystyle c\circ i\in\prod_{\alpha\in\mathcal{A}}U_\alpha$ from where it follows that $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$ Conversely, let $\Omega$ be a class of sets and let $i:\mathcal{A}\to\Omega$ be an indexing function. We may then index $\Omega$ by $\Omega=\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$. Then, by assumption $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$ Thus there exists some $\displaystyle \bold{x}\in\prod_{\alpha\in\mathcal{A}}U_\alpha$ Such that $\displaystyle \bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }\bold{x}(\alpha)\in U_\alpha$ Thus, we have that $\displaystyle \bold{x}\circ i^{-1}:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}}$ is a well-defined mapping with $\displaystyle \bold{x}\left(U_\alpha\right)\in U_\alpha$ For each $\alpha\in\mathcal{A}$. It follows that we have produced a choice function for $\Omega$ and the conclusion follows. $\blacksquare$ Remark: We have assumed the existence of a bijective indexing function $i:\mathcal{A}\to\Omega$, but this is either A) a matter for descriptive set theory or B) obvious since $\text{id}:\Omega\to\Omega$ satisfies the conditions. This depends on your level of rigor. 10. Problem: Let $A$ be a set; let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of spaces; and let $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of functions $f_\alpha:A\to X_\alpha$ a) Prove there is a unique coarsest topology $\mathfrak{J}$ on $A$ relative to whish each of the functions $f_\alpha$ is continuous. b) Let $\mathcal{S}_\beta=\left\{f_\beta^{-1}\left(U_\beta\right):U_\beta\text{ is ope}\text{n in}X_\beta\right\}$ and let $\displaystyle \mathcal{S}=\bigcup_{\alpha\in\mathcal{A}}$. Prove that $\mathcal{S}$ is a subbasis for $\mathfrak{J}$. c) Show that the map $g:Y\to A$ is continuous relative to $\mathfrak{J}$ if and only if each map $f_\alpha\circ g:Y\to X_\alpha$ is continuous. d) Let $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ be defined by the equation $f(x)=\left(f_\alpha(x)\right)_{\alpha\in\mathcal{A}}$ Let $Z$ denote the subspace of $f\left(A\right)$ of the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$. Prove taht the image under $f$ of each element of $\mathfrak{J}$ is an open set in $Z$. Proof: a) We first prove a lemma Lemma: Let $\mathfrak{J}$ be a topology on $A$, then all the mappings $f_\alpha:A\to X_\alpha$ are continuous if and only if $\mathcal{S}\subseteq\mathfrak{J}$ where $\mathcal{S}$ is defined in part b). Proof:Suppose that all the mappings $f_\alpha:A\to X_\alpha$ are continuous. Then, given any open set $U_\alpha\in X_\alpha$ we have that $f_\alpha$ is continuous and so $f_\alpha^{-1}\left(U_\alpha\right)$ is open and thus $f_{\alpha}^{-1}\left(U_\alpha\right)\in\mathfrak{J}$ from where it follows that $\mathcal{S}\subseteq\mathfrak{J}$. Conversely, suppose that $\mathcal{S}\subseteq\mathfrak{J}$. It suffices to prove that $f_\alpha:A\to X_\alpha$ for a fixed but arbitrary $\alpha\in\mathcal{A}$. So, to do this let $U$ be open in $X_\alpha$ then $f_{\alpha}^{-1}\left(U\right)\in\mathfrak{J}$ and thus by assumption $f_\alpha^{-1}\left(U\right)\in\mathfrak{J}$; but this precisely says that $f_\alpha^{-1}\left(U\right)$ is open in $A$. By prior comment the conclusion follows. $\blacksquare$ So, let $\mathcal{C}=\left\{\mathfrak{T}:\mathfrak{J}\text{ is a topology on }A\text{ and }\mathcal{S}\subseteq\mathfrak{J}\right\}$ and let $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in\mathcal{C}}\mathfrak{T}$ By previous problem $\mathfrak{J}$ is in fact a topology on $A$, and by our lemma we also know that all the mappings $f_\alpha:A\to X_\alpha$ are continuous since $\mathcal{S}\subseteq\mathfrak{J}$. To see that it’s the coarsest such topology let $\mathfrak{U}$ be a topology for which all of the $f_\alpha:A\to X_\alpha$ are continuous. Then, by the other part of our lemma we know that $\mathcal{S}\subseteq\mathfrak{U}$ and thus $\mathfrak{U}\in\mathcal{C}$. So, $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}\subseteq\mathfrak{U}$ And thus $\mathfrak{J}$ is coarser than $\mathfrak{U}$. The uniqueness is immediate. b) It follows from the previous problem that we must merely show that $\mathcal{S}$ is a subbasis for the topology $\mathfrak{J}$. The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons): Lemma: Let $X$ be a set and $\Omega$ be a subbasis for a topology on $X$. Then, the topology generated by $\Omega$ equals the intersection of all topologies which contain $\Omega$. Proof: Let $\mathcal{C}\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\Omega\subseteq\mathfrak{T}\right\}$ and $\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}$ Also, let $\mathfrak{G}$ be the topology generated by the subbasis $\Omega$. Clearly since $\Omega\subseteq\mathfrak{G}$ we have that $\mathfrak{J}\subseteq\mathfrak{G}$. Conversely, let $U\in\mathfrak{G}$. Then, by definition to show that $U\in\mathfrak{J}$ it suffices to show that $U\in\mathfrak{T}$ for a fixed but arbitrary $\mathfrak{T}\in\mathcal{C}$. To do this we first note that by definition that $\displaystyle U=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$ where each $U_\alpha=O_1\cap\cdots\cap O_{m_\alpha}$ for some $O_1,\cdots,O_{m_\alpha}\in\Omega$. Now, if $\mathfrak{T}\in\mathcal{C}$ we know (since $\Omega\subseteq\mathfrak{T}$) that $O_1,\cdots,O_{m_\alpha}\in\mathfrak{T}$ and thus $O_1\cap\cdots\cap O_{m_\alpha}=U_\alpha\in\mathfrak{T}$ for each $\alpha\in\mathcal{A}$. It follows that $U$ is the union of sets in $\mathfrak{T}$ and thus $U\in\mathfrak{T}$. It follows from previous comment that $\mathfrak{G}\subseteq\mathfrak{J}$. The conclusion follows. $\blacksquare$ The actual problem follows immediately from this. c) So, let $g:Y\to A$ be some mapping and suppose that $f_\alpha\circ g:Y\to X_\alpha$ is continuous for each $\alpha\in\mathcal{A}$. Then, given a subbasic open set $U$ in $A$ we have that $U=f_{\alpha_1}^{-1}\left(U_{\alpha_1}\right)\cap\cdots\cap f_{\alpha_n}^{-1}\left(U_{\alpha_n}\right)$ for some $\alpha_1,\cdots,\alpha_n$ and for some open sets $U_{\alpha_1},\cdots,U_{\alpha_n}$ in $X_{\alpha_1},\cdots,X_{\alpha_n}$ respectively. Thus $g^{-1}(U)$ may be written as $\displaystyle g^{-1}\left(\bigcup_{j=1}^{n}f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}g^{-1}\left(f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}\left(f_{\alpha_j}\circ g\right)^{-1}\left(U_{\alpha_j}\right)$ but since each $f_{\alpha_j}\circ g:Y\to X_{\alpha_j}$ we see that $g^{-1}\left(U\right)$ is the finite union of open sets in $Y$ and thus open in $Y$. It follows that $g$ is continuous. Conversely, suppose that $g$ is continuous then $f_\alpha\circ g:Y\to X_{\alpha}$ is continuous since it’s the composition of continuous maps. d) First note that $\displaystyle f^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}f\alpha^{-1}\left(U_\alpha\right)$ from where it follows that the initial topology under the class of maps $\{f_\alpha\}$ on $A$ is the same as the initial topology given by the single map $f$. So, in general we note that if $X$ is given the initial topology determined by $f:X\to Y$ then given an open set $f^{-1}(U)$ in $X$ we have that $f\left(f^{-1}(U)\right)=U\cap f(X)$ which is open in the subspace $f(X)$. June 9, 2010 ## Munkres Chapter 2 Section 19 (Part I) 1. Problem: Suppose that for each $\alpha\in\mathcal{A}$ the topology on $X_\alpha$ is given by a basis $\mathfrak{B}_\alpha$. The collection of all sets of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$ Such that $B_\alpha\in\mathfrak{B}_\alpha$ is a basis for $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ with the box topology, denote this collection by $\Omega_B$. Also, the collection $\Omega_P$ of all sets of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$ Where $B_\alpha\in\mathfrak{B}_\alpha$ for finitely many $\alpha$ and $B_\alpha=X_\alpha$ otherwise is a basis for the product topology on $X$. Proof: To prove the first part we let $U\subseteq X$ be open. Then, by construction of the box topology for each $(x_\alpha)_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ and $\displaystyle (x_\alpha)\in \prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$. So, then for each $x_\alpha$ we may find some $B_\alpha\in\mathfrak{B}_\alpha$ such that $x_\alpha\in B_\alpha\subseteq U_\alpha$ and thus $\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}B_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$ Noticing that $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha\in\Omega_B$ and every element of $\Omega_B$ is open finishes the argument. Next, we let $U\subseteq X$ be open with respect to the product topology. Once again for each $(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ for each $\alpha\in\mathcal{A}$ and $U_\alpha=X_\alpha$ for all but finitely many $\alpha$, call them $\alpha_1,\cdots,\alpha_m$. So, for each $\alpha_k,\text{ }k=1,\cdots,m$ we may find some $B_k\in\mathfrak{B}_k$ such that $x\in B_k\subseteq U_k$ and so $\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}V_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$ Where $\displaystyle V_\alpha=\begin{cases}B_k\quad\text{if}\quad \alpha=\alpha_k ,\text{ }k=1,\cdots,m\\ X_\alpha\quad\text{if}\quad x\ne \alpha_1,\cdots,\alpha_m\end{cases}$ Noting that $\displaystyle \prod_{\alpha\in\mathcal{A}}V_\alpha\in\Omega_P$ and $\Omega_P$ is a collection of open subsets of $X$ finishes the argument. $\blacksquare$ 2. Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of topological spaces such that $U_\alpha$ is a subspace of $X_\alpha$ for each $\alpha\in\mathcal{A}$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=Y$ is a subspace of $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$ if both are given the product or box topology. Proof: Let $\mathfrak{J}_S,\mathfrak{J}_P$ denote the topologies $Y$ inherits as a subspace of $X$ and as a product space respectively. Note that $\mathfrak{J}_S,\mathfrak{J}_P$ are generated by the bases $\mathfrak{B}_S=\left\{B\cap Y:B\in\mathfrak{B}\right\}$ (where $\mathfrak{B}$ is the basis on $X$ with the product topology), and $\displaystyle \mathfrak{B}_P=\left\{\prod_{\alpha\in\mathcal{A}}O_\alpha:O_\alpha\text{ is open in }U_\alpha,\text{ and equals }U_\alpha\text{ for all but finitely many }\alpha\right\}$ So, let $(x_\alpha)\in B$ where $B\in\mathfrak{B}_S$ then $\displaystyle B=Y\cap\prod_{\alpha\in\mathcal{A}}V_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)$ Where $V_\alpha$ is open in $X_\alpha$, and thus $V_\alpha\cap Y$ is open in $U_\alpha$. Also, since $V_\alpha=X_\alpha$ for all but finitely many $\alpha$ it follows that $V_\alpha\cap U_\alpha=U_\alpha$ for all but finitely many $\alpha$. And so $B\in\mathfrak{B}_P$. Similarly, if $B\in\mathfrak{B}_P$ then $\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha$ Where $O_\alpha$ is open in $U_\alpha$, but this means that $O_\alpha=V_\alpha\cap U_\alpha$ for some open set $V_\alpha$ in $X_\alpha$ and so $\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}V_\alpha\cap Y\in\mathfrak{B}_S$ From where it follows that $\mathfrak{B}_S,\mathfrak{B}_P$ and thus $\mathfrak{J}_S,\mathfrak{J}_P$ are equal. The case for the box topology is completely analgous. $\blacksquare$ 3. Problem: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of Hausdorff spaces, then $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ is Hausdorff with either the box or product topologies Proof: It suffices to prove this for the product topology since the box topology is finer. So, let $(x_\alpha),(y_\alpha)\in X$ be distinct. Then, $x_\beta\ne y_\beta$ for some $\beta\in\mathcal{A}$. Now, since $X_\beta$ is Hausdorff there exists disjoint open neighborhoods $U,V$ of $x_\beta,y_\beta$ respectively. So, $\pi_\beta^{-1}(U),\pi_\beta^{-1}(V)$ are disjoint open neighborhoods of $(x_\alpha),(y_\alpha)$ respectively. The conclusion follows. $\blacksquare$ 4. Problem: Prove that $\left(X_1\times\cdots\times X_{n-1}\right)\times X_n\overset{\text{def.}}{=}X\approx X_1\times\cdots\times X_n\overset{\text{def.}}{=}Y$. Proof: Define $\displaystyle \varphi:X\to Y:((x_1,\cdots,x_{n-1}),x_n)\mapsto (x_1,\cdots,x_n)$ Clearly this is continuous since $\pi_{\beta}\circ\varphi=\pi_\beta$ 5. Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it? Proof: If $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ where the latter is given the box topology then we have that each $\pi_\alpha$ is continuous and thus so is each $\pi_\alpha\circ f:A\to X_\alpha$. $\blacksquare$# 6. Problem: Let $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ be a sequence of points in the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$. Prove that $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ converges to $\bold{x}$ if and only if the sequences $\left\{\pi_\alpha(\bold{x}_n)\right\}_{n\in\mathbb{N}}$ coverge to $\pi_\alpha(\bold{x})$ for each $\alpha\in\mathcal{A}$. Is this fact true if one uses the box topology? Proof: Suppose that $U$ is a neighborhood of $\pi_{\alpha}(\bold{x})$ such that $\left(X_\alpha-U\right)\cap\left\{\bold{x}_n:n\in\mathbb{N}\right\}=K$ is infinite. Notice then that if $\pi_{\alpha}(\bold{x}_n)\in K$ that $\bold{x}_n\notin\pi_{\alpha}^{-1}\left(U\right)$ from where it follows that $\pi_{\alpha}^{-1}\left(U\right)$ is a neighborhood of $\bold{x}$ which does not contain all but finitely many values of $\left\{\bold{x}_n:n\in\mathbb{N}\right\}$ contradicting the fact that $\bold{x}_n\to\bold{x}$ in $X$. Conversely, suppose that $\pi_{\alpha}(\bold{x}_n)\to\pi_{\alpha}(\bold{x})$ for each $\alpha\in\mathcal{A}$ and let $\displaystyle \prod_{\alpha\in\mathcal{A}}U_{\alpha}$ be a basic open neighborhood of $\bold{x}$. Then, letting $\alpha_1,\cdots,\alpha_m$ be the finitely many indices such that $U_{\alpha_k}\ne X_k$. Since each $\pi_{\alpha_k}(\bold{x}_n)\to\pi_{\alpha_k}(\bold{x})$ there exists some $n_\ell\in\mathbb{N}$ such that $n_\ell\leqslant n\implies \pi_{\alpha_k}(\bold{x}_k)\in U_k$. So, let $N=\max\{n_1,\cdots,n_k\}$. Now, note that if $\displaystyle \bold{x}_n\notin\prod_{\alpha\in\mathcal{A}}U_\alpha$ then $\pi_{\alpha}(\bold{x}_n)\notin U_\alpha$ for some $\alpha\in\mathcal{A}$. But, since clearly $\pi_{\alpha}(\bold{x}_n)\in X_\alpha$ we must have that $\pi_{\alpha}(\bold{x}_n)\notin U_{\alpha_1}\cup\cdots\cup U_{\alpha_m}$ and thus $n\leqslant N$. It follows that for every $N\leqslant n$ we have that $\displaystyle \bold{x}_n\in\prod_{\alpha\in\mathcal{A}}U_\alpha$. Then, since every neighborhood of $\bold{x}$ contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows. Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider $\displaystyle \bold{x}_n=\prod_{m\in\mathbb{N}}\left\{\frac{1}{n}\right\}$. Clearly each coordinate converges to zero, but $\displaystyle \prod_{m\in\mathbb{N}}\left(\frac{-1}{m},\frac{1}{m}\right)=U$ is a neighborhood of $\bold{0}$ in the product topology. But, if one claimed that for every $n\geqslant N$ (for some $N\in\mathbb{N}$ that $\bold{x}_n\in U$ they’d be wrong. To see this merely note $\displaystyle \frac{1}{N}\notin\left(\frac{-1}{N+1},\frac{1}{N+1}\right)$ and so $\pi_{N+1}\left(\bold{x}_N\right)\notin\pi_{N+1}(U)$ and thus $\bold{x}_{N}\notin U$. 7. Problem: Let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all eventually zero sequences. What is $\overline{\mathbb{R}^{\infty}}$ in the box and product topology? Proof: We claim that in the product topology $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}$. To see this let $\displaystyle \prod_{n\in\mathbb{N}}U_n$ be a basic non-empty open set in $\mathbb{R}^{\omega}$ with the product topology. Since we are working with the product topology we know there are finitely many indices $n_1,\cdots,n_m$ such that $U_{n_k}\ne \mathbb{R}$. So, for each $n_1,\cdots,n_m$ select some $x_{n_k}\in U_{n_k}$ and consider $(x_n)_{n\in\mathbb{N}}$ where $\displaystyle \begin{cases} x_{n_k}\quad\text{if}\quad n=n_1,\cdots,n_m \\ 0 \text{ } \quad\text{if}\quad \text{otherwise}\end{cases}$ Clearly then $\displaystyle (x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}}U_n\cap\mathbb{R}^{\infty}$ and thus every non-empty open set in $\mathbb{R}^{\omega}$ intersects $\mathbb{R}^{\infty}$ and the conclusion follows. Now, we claim that with the box topology that $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$. To see this let $(x_n)_{n\in\mathbb{N}}\notin\mathbb{R}^{\infty}$. Then, there exists some subsequence $\{x_{\varphi(n)}\}$ of the sequence $\{x_n\}$ which is non-zero. For each $\varphi(n)$ form an interval $I_{\varphi(n)}$ such that $0\notin I_{\varphi(n)}$. Then, consider $\displaystyle \prod_{n\in\mathbb{N}}U_n,\text{ }U_n=\begin{cases} I_{\varphi(m)}\quad\text{if}\quad n=\varphi(m)\\ \mathbb{R}\quad\text{ }\quad\text{if}\quad \text{otherwise}\end{cases}$ Clearly then $\displaystyle \prod_{n\in\mathbb{N}}U_n$ is a neighborhood of $(x_n)_{n\in\mathbb{N}}$ and since each clearly has an infinite subsequence of non-zero values it is disjoint from $\mathbb{R}^{\infty}$. It follows that in $\mathbb{R}^{\omega}$ with the box topology that $\mathbb{R}^{\infty}$ is closed and thus $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$ as desired. $\blacksquare$ 8. Problem: Given sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ of real numbers with $a_n>0$ define $\varphi:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega}$ by the equation $\varphi:(x_n)_{n\in\mathbb{N}}\mapsto \left(a_n x_n+b_n\right)_{n\in\mathbb{N}}$ Show that if $\mathbb{R}^{\omega}$ is given the product topology that $\varphi$ is a homeomorphism. What happens if $\mathbb{R}^{\omega}$ is given the box topology? Proof: Let us first prove that $\varphi$ is a bijection. To do this we prove something more general… Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of untopologized sets and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a collection of bijections $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ we have that $\varphi$ is a bijection. Proof: To prove injectivity we note that if $\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\varphi\left((y_\alpha)_{\alpha\in\mathcal{A}}\right)$ Then, $\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=\left(f_\alpha(y_\alpha)\right)_{\alpha\in\mathcal{A}}$ And by definition of an $\alpha$-tuple this implies that $f_\alpha(x_\alpha)=f_\alpha(y_\alpha)$ for each $\alpha\in\mathcal{A}$. But, since each $f_\alpha:X_\alpha\to Y_\alpha$ is injective it follows that $x_\alpha=y_\alpha$ For each $\alpha\in\mathcal{A}$. Thus, $(x_\alpha)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$ as desired. To prove surjectivity we let $\displaystyle (y_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}Y_\alpha$ be arbitrary. We then note that for each fixed $\alpha\in\mathcal{A}$ we have there is some $x_\alpha\in X_\alpha$ such that $f_\alpha(x_\alpha)=y_\alpha$. So, if $\displaystyle (x_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}X_\alpha$ is the corresponding $\alpha$-tuple of these values we have that $\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$ from where surjectivity follows. Combining these two shows that $\varphi$ is indeed a bijection. $\blacksquare$ Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a corresponding class of continuous functions such that $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topologies the mapping $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ is continuous. Proof: Since the codomain is a product space it suffices to show that $\displaystyle \pi_\beta\circ\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to Y_{\beta}$ is continuous for each $\beta\in\mathcal{A}$. We claim though that the diagram  commutes where $\pi^Y_\beta$ and $\pi^X_\beta$ denote the canonical projections from $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ to $Y_\beta$ and $X_\beta$ respectively. To see this we merely note that $\displaystyle \pi^Y_\beta\left(\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=\pi^Y_\beta\left(\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}\right)=f_{\beta}(x_\beta)$ and $f_\beta\left(\pi^X_\beta\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=f_\beta\left(x_\beta\right)$ which confirms the commutativity of the diagram. But, the conclusion follows since $f_\beta\circ\pi_\beta$ is the composition of two continuous maps (the projection being continuous since $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ is a product space). The lemma follows by previous comment. $\blacksquare$ We come to our last lemma before the actual conclusion of the problem. Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a set of homeomorphisms with $f_\alpha:X_\alpha\to Y_\alpha$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$ is a homeomorphism if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topology. Proof: Our last two lemmas show that $\varphi$ is bijective and continuous. To prove that it’s inverse is continuous we note that $\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}\right)\circ\varphi\right)\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\prod_{\alpha\in\mathcal{A}}\left\{f_\alpha^{-1}\left(f_\alpha(x_\alpha)\right)\right\}=(x_\alpha)_{\alpha\in\mathcal{A}}$ And similarly for the other side. Thus, $\displaystyle \varphi^{-1}=\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}$ Which is continuous since each $f_{\alpha}^{-1}:Y_\alpha\to X_\alpha$ is continuous and appealing to our last lemma again. Thus, $\varphi$ is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows. $\blacksquare$ Thus, getting back to the actual problem we note that if we denote $T_n:\mathbb{R}\to\mathbb{R}:x\mapsto a_nx+b_n$ that each $T_n$ is a homeomorphism. Thus, since it is easy to see that $\displaystyle \varphi=\prod_{n\in\mathbb{N}}T_n$ we may conclude by our last lemma (since we are assuming that we are giving $\mathbb{R}^{\omega}$ in both the domain and codomain the product topology) that $\varphi$ is a homeomorphism. This is also continuous if we give $\mathbb{R}^{\omega}$ the box topology. To see this we merely need to note that $\displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}f_\alpha^{-1}\left(U_\alpha\right)$ and thus if all of the $U_\alpha$ are open then so are (since each $f_\alpha$ is continuous) is each $f_\alpha^{-1}(U_\alpha)$ and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function. $\blacksquare$ June 7, 2010 ## Topological groups (Direct Product and Product Spaces) Up until now we’ve discussed how the subspace topology reacts with subgroups and how the quotient topology interacts with the quotient group, it seems like a natural progression to then discuss how the product topology reacts with the direct product. So, as in the past we firstly need to verify that the two do agree, namely: Theorem: Let $G,G'$ be topological groups. Then, $G\times G'$ is a topological group with the direct product group structure and product topology. Proof: Clearly $G_1\times\cdots\times G_n$ is a set with both a group theoretic and topological structure and so it remains to show that the map $\alpha:(G\times \cdots\times G_n)\times (G\times \cdots\times G_n)\to G\times \cdots\times G_n$ given by $\alpha:\left((g_1,\cdots,g_n),(g'_1,\cdots,g'_n)\right)\mapsto (g_1{g'_1}^{-1},\cdots,g_n{g'_n}^{-1})$ is continuous. But, note that $(G\times \cdots\times G_n)\times (G\times \cdots\times G_n)\approx (G\times G)\times\cdots\times (G_n\times G_n)$ And that $\alpha_G\times\cdots\times \alpha_{G_n}:(G\times G)\times\cdots\times (G_n\times G_n)$ given by $\alpha:\left((g_1,g'_1),\cdots,(g_n,g'_n)\right)\mapsto\left(g_1{g'_1}^{-1},\cdots,g_n{g'_n}^{-1}\right)$ being the product of continuous maps is continuous. The conclusion immediately follows. $\blacksquare$ There are canonical topological epimorphisms from $G_1\times\cdots\times G_n$ into $G_k$. Namely: Theorem: Let $\pi_k:G_1\times G_n\to G_k:(g_1,\cdots,g_k,\cdots,g_n)\mapsto g_k$ Then, $\pi_k$ is an open topological epimorphism. Proof: The fact that it’s open, surjective, and continuous follows since topologically $\pi_k$ is merely the canonical projection of a product space onto it’s $k$th coordinate. Thus, it remains to show that it’s a homomorphism. But, this is a routine calculation. $\blacksquare$ Theorem: Let $\pi_k$ be as above. Then, $\ker\pi_k=G_1\times\cdots\times\underbrace{\{e\}}_{k}\times\cdots\times G_n$ from where it follows from a previous theorem that $\left(G_1\times\cdots\times G_n\right)/\left(G_1\times\cdots\times\{e\}\times\cdots\times\right)\overset{\text{T.G.}}{\cong}G_k$. $\blacksquare$ Now, so we can use it we should mention in passing that: Theorem: $G_1\times\cdots\times G_n\overset{\text{T.G.}}{\cong}G_{\sigma(1)}\times\cdots\times G_{\sigma(n)}$ where $\sigma:\{1,\cdots,n\}\to\{1,\cdots,n\}$ is any bijeciton. Proof: Obvious. $\blacksquare$ In fact, that is all I wanted to mention about them. Unfortunately the interesting stuff about the direct product of topological groups that is interesting isn’t “accessible” and would take too much time to really discuss. Thus, for once I have a post which is less than one-thousand words. Enjoy. April 29, 2010 ## The dreaded product topology (Pt. III Tychonoff’s theorem and odds and ends) In this post we will discuss one of the most important in point-set topology, Tychonoff’s theorem, and prove up some miscellaneous facts that are either interesting or useful. Theorem (Tychonoff): Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty class of non-empty compact topological spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is compact under the product topology. Proof: It follows from an earlier theorem that we must merely show that every class of closed sets in the defining closed subbase with the F.I.P have non-empty intersection. So, let $\mathcal{S}=\left\{S_k\right\}_{k\in\mathcal{K}}$ be such a class. Consider some arbitrary $j\in\mathcal{j}$, since $S_k$ is subbasic closed for ever $S_k\in\mathcal{S}$ we know that $\pi_j(S_k)$ is closed, and so $\left\{\pi_j(S_k)\right\}_{j\in\mathcal{J}}$ is a class of closed subsets of $X_j$, and noticing that $\pi_{j}(S_{k_1})\cap\cdots\cap\pi_j(S_{k_n})\supseteq\pi_j\left(S_{k_1}\cap\cdots\cap S_{k_n}\right)\supseteq\varnothing$ and so by assumption we have that $\displaystyle \bigcap_{k\in\mathcal{K}}\pi_j(S_k)\ne \varnothing$. In other words, we have there exists some $\displaystyle x_j\in\bigcap_{k\in\mathcal{K}}\pi_j(S_k)$ and so every element of $\mathcal{S}$ contains an element which has $x$ as it’s $j$th coordinate. Doing this for each $j\in\mathcal{J}$ produces a set $\left\{x_j\right\}_{j\in\mathcal{J}}$ such that $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\bigcap_{k\in\mathcal{K}}S_k$. In light of our initial comments the conclusion follows. $\blacksquare$ From this we can prove a nice little result that is used incessantly in multivariate analysis. But, before we continue we need a little terminology n-cell: An $n$-cell is a multi-dimensional generalization of a closed interval in $\mathbb{R}^n$ . In other words, we call $E$ an $n$-cell if $\pi_k(E)=[a,b]$ for every $1\leqslant k\leqslant n$. In other words, $E$ is an $n$-cell if there exists some $[a,b]\subseteq\mathbb{R}$ such that $E=\underbrace{[a,b]\times\cdots\times[a,b]}_{n\text{ times}}=[a,b]^n$. Theorem (Generalized Heine-Borel): Every closed and bounded subspace of $\mathbb{R}^n$ is compact. Proof: Let $E$ be such a set. Since $E$ is bounded we know that $E$ is contained in some $n$-cell $[a,b]^n$. So, if we prove that every $n$-cell is compact we are done (since a closed subspace of a compact space is compact). But, we know that $[a,b]^n=[a,b]\times\cdots\times[a,b]$ and since $[a,b]\subseteq\mathbb{R}$ is compact we have by Tychonoff’s theorem that the product of compact spaces is compact. The conclusion follows. $\blacksquare$ Be careful about taking this and attempting to try to make a generalization of the extreme value theorem, namely “$If $f:X\mapsto\mathbb{R}^n$ is continuous and $X$ compact then $\sup\text{ }f\left(X\right),\inf\text{ }f\left(X\right)\in f\left(X\right)$“. It fails for an easily over-looked and decidedly non-topological property of $\mathbb{R}^n,text{ }n\geqslant 2$. Particularly, we have that $\mathbb{R}$ has a canonical ordering (the one we know and love), but what is the ordering on $\mathbb{R}^n$? No ordering no concept of infimums and supremums.

We now take a look at some of the odds and ends that were previously mentioned. It should probably be mentioned for those that are reading this and already know topology that, for now, we omit mention of the majority (save Hausdorfness AKA $T_2$) of the separability axioms and things like local compacntess, connectedness, paracompactness, and path-connectedness. We shall mention them in due time, from where we will discuss which are invariant under product.

Theorem: Let $X_1,\cdots,X_n$ be a finite number of discrete spaces, then the product set $X=X_1\times\cdots\times X_n$ is the same topological space under both the product and discrete topology.

Proof: Let $\mathcal{J}$ be the product topology on $X$ and, of course, let $\wp\left(X\right)$ be the discrete topology (I use the $\wp$ for power set). It is clear that $\mathcal{J}\subseteq\wp\left(X\right)$ and so we must merely show the reverse inclusion to finish.

So, let $E\in\wp\left(X\right)$ then $E=U_1\times\cdots\times U_n$ for some subsets $U_k\subseteq X_k$, but since $X_k$ is discrete we know that each $U_k$ is open. Thus, $E$ is the finite product of open sets and thus open in the product (box) topology. The conclusion follows. $\blacksquare$

The fact that I noted that product topology and box topology were the same here was meaningful. We know that the two coincide  for the product of finitely many spaces, but the next remark shows (once again) how they differ for infinite.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of discrete spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ under the box topology coincides with $X$ under the discrete topology. The same cannot be said about the product topology.

Proof: Follows the exact same reasoning as in the last theorem.

To see why the infinite product of discrete spaces need not be discrete under the product topology, one need only consider $\displaystyle X=\prod_{j=1}^{\infty}X_j$ where $X_j=[0,1]$ under the discrete topology. Clearly then $(0,1)\times\cdots\times(0,1)\times\cdots$ is open in $X$ under the discrete topology but NOT the product topology. $\blacksquare$

The last thing we will talk about is the product of maps.

Product of maps: Let $f:X\mapsto Y$ and $f':X'\mapsto Y'$ be two maps, then define the product of the maps to be $f\times f':X\times X'\mapsto Y\times Y'$ by $(x,x')\mapsto (f(x),f'(x'))$.

We extend this definition to an arbitrary number of topological spaces as follows: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ and $\left\{Y_j\right\}_{j\in\mathcal{J}}$ be two sets of topological spaces and let $\left\{f_j\right\}_{j\in\mathcal{J}}$ be a class of mappings such that $f_j:X_j\mapsto Y_j$ (this can obviously be formulated in the guise of ordered triples as well), we then define $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{j}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$  by $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\mapsto\prod_{j\in\mathcal{j}}\{f_j(x_j)\}$

The reader may recall that this concept was used to prove that if $f:X\mapsto Y$ is continuous and $Y$ Hausdorff then $\Gamma_f$ is a closed subset of $X\times Y$.

We begin by proving the product of continuous maps is continuous. This is the first of a sequence of three proofs which will lead to a very satisfying corollary.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ and $\left\{Y_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of topological spaces and let $\left\{f_j\right\}_{j\in\mathcal{J}}$ be a corresponding set of continuous mappings such that $f_j:X_j\mapsto Y_j$. Then, $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is continuous, where $\displaystyle \prod_{j\in\mathcal{J}}X_j,\prod_{j\in\mathcal{J}}Y_j$ are under the product topology.

Proof:

Lemma: $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j^{-1}\left(O_j\right)$

Proof: Let $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\left( \prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)$ then, $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}\in\prod_{j\in\mathcal{J}}O_j$ which clearly means

that $f_j(x_j)\in O_j,\text{ }\forall j\in\mathcal{J}$ and so $x_j \in f^{-1}(O_j)$ and thus

$\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$

Conversely, let $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$. Then, $x_j\in f^{-1}\left(O_j\right)\implies f(x_j)\in O_j,\text{ }\forall j\in\mathcal{J}$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)\in \prod_{j\in\mathcal{J}}O_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)$. The conclusion follows. $\blacksquare$

Now, as was previously proven we need only prove that the inverse image of each open basic set is open. So, let $B\in\mathfrak{B}$ (the defining open base for $\displaystyle \prod_{j\in\mathcal{J}}Y_j$). Then, $\displaystyle B=\prod_{j\in\mathcal{J}}O_j$ where $O_j=Y_j$for all but finitely many $j$, and thus $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(B\right)=\left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$. But, we see that since $O_j$ is an open set which equals $Y_j$ for all but finitely many $j$‘s that $f^{-1}(O_j)$ is going to be an open set (since each $f_j:X_j\mapsto Y_j$ is continuous) which equals $X_j$ for all but finitely many $j$‘s. The conclusion follows. $\blacksquare$

Theorem: Let everything be as in the previous theorem, except instead of each $f_j:X_j\mapsto Y_j$ being continuous, let it be bijective. Then $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is bijective.

Proof:

Injectivity: Suppose that $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{y_j\}\right)$. Then, by definition $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{f_j(y_j)\}$ and since two functions (remember we think of arbitrary tuples as functions) are equal if and only if they take the same value at every point we see that $f_j(x_j)=f_j(y_j),\text{ }\forall j\in\mathcal{J}$ and since each $f_j$ is bijective (and thus injective) it follows that $x_j=y_j,\text{ }\forall j\in\mathcal{J}$, from where injectivity follows.

Surjectivity: Let $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}Y_j$ be arbitrary. For each $j\in\mathcal{J}$ we have that there exists some $x_j\in X_j$ such that $f_j(x_j)=y_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}$ from where surjectivity immediately follows. $\blacksquare$

Theorem: Let everything be as before, except now instead of continuous or bijective assume that each $f_j:X_j\mapsto Y_j$ is surjective and open. Then, $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is open.

Proof:

Lemma: $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j(O_j)$.

Proof: Let $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)$ then $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}x_j\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}$ for some $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}(O_J)$. Clearly then we have that $x_j\in O_j$ and so $f_j(x_j)\in O_j$ and so $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j(O_j)$.

Conversely, let $\displaystyle\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}\{f_j(O_j)\}$ then $y_j=f_j(x_j),\text{ }\forall j\in\mathcal{J}$ for some $x_j\in O_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)$

The conclusion follows. $\blacksquare$.

Now, let $E$ be open in $\displaystyle \prod_{j\in\mathcal{J}}X_j$ then $\displaystyle E=\prod_{j\in\mathcal{J}}O_j$ where $O_j=X_j$ for all but finitely many $j$ it follows then that $f_j(O_j)$ is an open set which equals $Y_j$ for all but finitely many $j$ (since each $f_j:X_j\mapsto Y_j$ is surjective and open). Thus $\displaystyle \prod_{j\in\mathcal{J}}f_j(E)=\prod_{j\in\mathcal{J}}f_J\left(\prod_{j\in\mathcal{J}}\right)=\prod_{j\in\mathcal{J}}f_j(O_j)$ which by prior comment is the product of open sets in $\displaystyle \prod_{j\in\mathcal{J}}Y_j$ which equals the full space for all but finitely many $j$. It follows that $f(E)$ is open in $\displaystyle \prod_{j\in\mathcal{J}}Y_j$ under the product topology. $\blacksquare$
Theorem: Combining these three we see that if $X_j\approx Y_j$ for all $j\in\mathcal{J}$, then there exists some homeomorphism (a bijective continuous open map)  $f_j:X_j\mapsto Y_j$ and so $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is a bijective continuous open map and thus a homeomorphism. It follows that $\displaystyle \prod_{j\in\mathcal{J}}X_j\approx\prod_{j\in\mathcal{J}}Y_j$ where each are under the product topology.

February 21, 2010

## The dreadful product topology (Pt. II)

In the previous post we gave the definitions of the product topology and some nice little tools about it. Here, we give some theorems regarding what properties of topological spaces are invariant under the formation of the product topology. Let us start out nice and easy

Theorem: Let $X_1,\cdots,X_n$ be a finite number of topological spaces with corresponding open bases $\mathfrak{B}_1,\cdots,\mathfrak{B}_n$. Then, $\mathfrak{M}=\left\{B_1\times\cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant m\right\}$ is an open base for $X_1\times\cdots\times X_n$ under the product topology.

Proof: Let $O$ be open in $X$, and let $o\in O$. There exists some $B\in\mathfrak{B}$ (the defining open base) such that $o\in B\subseteq O$. But, by theorem 1 in the last post we know that $\pi_k(B)$ is an open set containing $\pi_k(o)$ for $1\leqslant k\leqslant n$ and since $\mathfrak{B}_k$ is an open base for $X_k$ we know there exists some $J_k\in\mathfrak{B}_k$ such that $\pi_k(o)\in J_k\subseteq \pi_k(B)$. Clearly then $o\in J_1\times J_n\subseteq B$ and since $J_1\times J_n\in\mathfrak{M}$ and $o,O$ were arbitrary the conclusion follows. $\blacksquare$

We know use this to prove a nice little theorem.

Theorem: Let $X_1,\cdots,X_n$ be a finite number of second countable spaces, then $X=X_1\times\cdots\times X_n$ under the product topology is second countable.

Proof: Let $\mathfrak{M}$ be as above. Since $\mathfrak{M}$ is an open base for $X$ is remains to prove that it is countable. Let $\eta:\mathfrak{M}\mapsto\mathfrak{B}_1\times\cdots\times\mathfrak{B}_n$ be given by $B_1\times\cdots\times B_n\mapsto \left(B_1,\cdots,B_n\right)$. It is clear that this is a bijection. Recalling that the finite product of countable sets is countable finishes the argument. $\blacksquare$

It turns out that the above is a little more restrictive than is necessary as the following theorem illustrates.

Theorem: Let $\left\{X_n\right\}_{n\in\mathbb{N}}$ be a sequence of second countable spaces, then $\displaystyle X=\prod_{n=1}^{\infty}X_n$ is second countable under the product topology.

Proof: Let $\left\{\mathfrak{B}_n\right\}$ be the corresponding countable open bases and define $\displaystyle \mathcal{B}_m=\left\{E:E=\prod_{j=1}^{\infty}G_j\right\}$ where $\pi_k(G_j)\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant m$ and $\pi_k(G_j)=X_j,\text{ }m.

It is relatively easy to prove that $\eta:\mathcal{B}_m\mapsto\mathfrak{B}_1\times\cdots\times\mathfrak{B}_m$ given by $\displaystyle B_1\times\cdots\times B_m\times\prod_{j=m+1}^{\infty}X_{j+1}\mapsto \left(B_1,\cdots,B_m\right)$  is a bijection. It follows then, since the finite product of countable sets is countable, that $\mathcal{B}_m$ is countable.

So, now define $\displaystyle \mathfrak{M}=\bigcup_{j=1}^{\infty}\mathcal{B}_j$. It is clear that since $\mathfrak{M}$ is the countable union of countable sets that it is countable. So now, it remains to prove that $\mathfrak{M}$ is in fact an open base for $X$

So, let $O$ be open in $X$ and let $o\in O$. There exists some $B$ in the defining open base such that $o\in B\subseteq O$. Considering the form in which basic open sets in the product topology take we know that there is a finite set of indices $\left\{j_1,\cdots,j_n\right\}$ such that $\pi_{j_k}(B)\ne X_k$. So, let $m=\max\{j_1,\cdots,j_n\}$. Clearly we have that $\pi_\ell(O)$ is a neighborhood of $\pi_k(o)$ for $1\leqslant k\leqslant m$ and since $\mathfrak{B}_k$ is an open base there exists some $B_k$ such that $\pi_k(o)\in B_k\subseteq\pi_k(O)$. So, let $U=B_1\times\cdots\times B_m\times$$\displaystyle \prod_{j=m+1}^{\infty}X_j$. Then, it is evident that $o\in U\subseteq B\subseteq O$. Noting that $U\in\mathfrak{M}$ finishes the argument. $\blacksquare$.

Note the necessity for finiteness again.

Theorem: Let $X_1,\cdots,X_n$ be a finite number of toplogical spaces and let $\mathfrak{D}_1,\cdots,\mathfrak{D}_n$ be a number of corresponding dense sets. Then, $\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n$ is dense in $X=X_1\times\cdots\times X_n$ under the product topology.

Proof: Let $x\in X$ be arbitrary and let $N$ be any neighborhood of $x$. We then have by theorem 1 in the last post that $\pi_k(N)$ is a neighborhood of $\pi_k(x)$ for $1\leqslant k\leqslant n$ and since $\mathfrak{D}_k$ is dense in $X_k$ there exists some $\mathfrak{d}_k\in\mathfrak{D}_k$ such that $\mathfrak{d}_k\in\pi_k(N)$. Clearly then $\left(\mathfrak{d}_1,\cdots,\mathfrak{d}_n\right)\in N$, and clearly $\left(\mathfrak{d}_1,\cdots,\mathfrak{d}_n\right)\in\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n$. Noting the arbitrariness of $x, N$ finishes the argument. $\blacksquare$

From this we can derive the obvious consequence.

Theorem: Let $X_1,\cdots,X_n$ be finite number of separable spaces and let $X=X_1\times\cdots\times X_n$ under the product topology, then $X$ is separable.

Proof: Since each $X_k$ is separable it has a countable dense subset $\mathfrak{D}_k$ and by the above theorem we see that $\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n$ is dense in $X$, but since the finite product of countable sets is countable the conclusion follows. $\blacksquare$

As in the case of second countable spaces we may extend this to the countable case

Theorem: Let $\left\{X_n\right\}_{n\in\mathbb{N}}$ be sequence of separable spaces, then $\displaystyle X=\prod_{j=1}^{\infty}X_j$ is separable.

Proof: Since each $X_n$ is separable it has a countable dense subset $\mathfrak{D}_n$ we may extract an arbitrary but fixed sequence from them as $\left\{d_n\right\}_{n\in\mathbb{N}}$ where $d_n\in\mathfrak{D}_n$. We now define $\displaystyle \mathcal{D}_m=\prod_{j=1}^{\infty}G_j$ where $G_j=\begin{cases} \mathfrak{D}_j & \mbox{if} \quad 1\leqslant j\leqslant m \\ \{d_j\} & \mbox{if} \quad j>m\end{cases}$

Clearly we have that $\eta:\mathcal{D}_m\mapsto\mathfrak{D}_1\times\cdots\times\mathfrak{D}_m$ by $D\mapsto \left\{\pi_1(D)\right\}\times\cdots\times \left\{\pi_m(D)\right\}$ is a bijection and since the finite product of countable sets is countable it follows that $\mathcal{D}_m$ is countable. So, let $\displaystyle \mathcal{D}=\bigcup_{m=1}^{\infty}\mathcal{D}_m$ being the countable union of countable sets is countable.

From the previous paragraph it is clear that we must merely show that $\mathcal{D}$ is dense in $X$. To do this let $x\in X$ be arbitrary and $N$ any neighborhood of $X$. There exists some $B\in\mathfrak{B}$ (the defining open base) such that $x\in B\subseteq N$. But, we know that $\pi_{k}(B)\ne X_k$ for finitely man indices, say $k_1,\cdots,k_m$. Let $r=\max\{k_1,\cdots,k_m\}$. We know then that $\pi_\ell(B)$ is a neighborhood of $\pi_\ell(x)$ for every $1\leqslant \ell\leqslant r$ and since $\mathfrak{D}_\ell$ is dense in $X_\ell$ there exists some $\mathfrak{d}_\ell$ such that $\mathfrak{d}_\ell\in\pi_\ell(B)$. Clearly then, we have that $\displaystyle D=\prod_{j=1}^{\infty}\{g_j\}$ where $g_j=\begin{cases} \mathfrak{d}_j & \mbox{if} \quad 1\leqslant j\leqslant r\\ d_j & \mbox{if} \quad j>r\end{cases}$ is in $B$ and thus $N$. But, $D\in\mathcal{D}$ and the conclusion follows. $\blacksquare$

We now discuss the product of Hausdorff spaces.

Theorem: $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of Hausdorff spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ under the product topology is Hausdorff.

Proof: Let $\bold{x},\bold{y}\in X$ such that $\bold{x}\ne \bold{y}$ then $\pi_k(\bold{x})\ne\pi_k(\bold{x})$ for some $k\in\mathcal{J}$ and since $X_k$ is Hausdorff we know that there exists open sets $U,V$ such latex $\pi_k(\bold{x})\in U,\pi_k(\bold{y})\in V$ and $U\cap V$. So define $\displaystyle E=\prod_{j\in\mathcal{J}}G_j$ where $G_j=\begin{cases} U & \mbox{if}\quad j=k \\ X_j & \mbox{if}\quad k\ne j\end{cases}$ and define $\displaystyle F=\prod_{j\in\mathcal{J}}H_j$ where $H_j=\begin{cases} V & \mbox{if}\quad j=k \\ X_j & \mbox{if} \quad x\ne j\end{cases}$ clearly we have that $\bold{x}\in E$ and $\bold{y}\in F$ and since $\pi_k(F)\cap\pi_k(E)=\varnothing$ we must have $F\cap E=\varnothing$. The conclusion follows. $\blacksquare$

February 20, 2010

NOTE: All sets and spaces in the following discussion will be assumed to be, without loss of generality, non-empty. Also, if not stated it is assumed that $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ under the product topology where $\left\{X_j\right\}_{j\in\mathcal{J}}$ is a non-empty collection of topological spaces.

To continue any further we mus make an, admittedly not brief, sojourn into the concept of product topologies.  Now, this isn’t a text-book so I’ll assume that the reader is familiar with the basic concepts of Cartesian products and projections (denote $\pi$). With this in mind we first make a definition for the arbitrary Cartesian product of sets:

Arbitrary Cartesian Product: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary class of non-empty sets, then we define $\displaystyle \prod_{j\in\mathcal{J}}X_j=\left\{\bold{x}:\mathcal{J}\mapsto\bigcup_{j\in\mathcal{J}}X_j \text{ }:\text{ } \bold{x}(j)\in X_j,\text{ }\forall j\in\mathcal{J}\right\}$.

Now, if $\left\{X_j\right\}_{j\in\mathcal{J}}$ is a non-empty class of topological spaces we define the product topology on $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ as the topology generated by the subbase $\mathfrak{S}=\left\{\pi_j^{-1}(G): G \in\mathfrak{J}_j\right\}$ (where $\mathfrak{J}_j$ is the topology on $X_j$,) in the sense that it is the set of all finite intersections of elements of $\mathfrak{S}$ form an open base $\mathfrak{B}$ for $X$, and thus the topology on $X$ is made up of all arbitrary unions of elements of this open base. We call $\mathfrak{S}$ in the above discussion the defining open subbase for $X$ and $\mathfrak{B}$ the defining open base for $X$ under the product topology.

So what do elements of $\mathfrak{S}$ look like? Well, let $G$ be an open set in $X_j$ then $\pi_j^{-1}(G)$ is (in words) all $\bold{x}\in X$ such that $\pi_j(\bold(x))\in G$. It is clear that the only limitation then that needs to be made on $\bold{x}$ is that it’s $j$th coordinate is in $G$. From this it is clear that $\displaystyle \pi_j^{-1}(G)=\prod_{\ell\in\mathcal{J}}E_\ell$ where $\displaystyle \begin{cases} X_j & \mbox{if} \quad \ell\ne j\\ G & \mbox{if} \quad \ell=j\end{cases}$. Thus, $\mathfrak{S}$ is made up of all sets of the form $\displaystyle \prod_{j\in\mathcal{J}}E_j$ where $E_j=X_j$ for all but one $j$

Example: If $X_j=[0,1],\text{ }j\in[0.1]$ under the usual topology then $\displaystyle (0,1)\times\prod_{j\in(0,1]}X_j\in\mathfrak{S}$ whereas $\displaystyle (0,1)\times\prod_{j\in(0,1)}X_j\times(0,1)\notin\mathfrak{S}$

So, this being said we see that the elements of $\mathfrak{B}$ (being the finite intersection of elements of $\mathfrak{S}$) are going to be the product of open sets $G_j\subseteq X_j$ such that $G_j=X_j$ for all but finitely many $j$.

So, let’s start proving some theorems!

Theorem 1: All projections $\pi_j:X\mapsto X_j$ are open mappings.

Proof: Let $E\subseteq X$ be open then $\displaystyle E=\bigcup_{k\in\mathcal{K}}B_k$ where $\left\{B_k\right\}_{k\in\mathcal{K}}\subseteq\mathfrak{B}$, and so $\displaystyle \pi_j(E)=\pi_j\left(\bigcup_{k\in\mathcal{K}}B_k\right)=\bigcup_{k\in\mathcal{K}}\pi_j\left(B_k\right)$ and since the $j$th coordinate of every basic open set (whether equal to the full space or not) is an open set in $X_j$ and so $\pi_j(E)$ is the union of open sets in$X_j$ and thus open. $\blacksquare$

Corollary 1: If $N$ is a neighborhood of $\bold{x}\in X$ then $\pi_j(N)$ is a neighborhood of $\pi_j(\bold{x})$

Remark: Projections need not be closed mappings. Consider the set $E=\left\{(x,y):\text{ }xy=1,\right\}$ as a subset of $\mathbb{R}^2$ with the product topology. It is relatively easy $E$ is closed, but $\pi_1(E)=\pi_2(E)=\mathbb{R}-\{0\}$ which is open.

One might ask “Why not just make the open base for a topology on $X$ the set of all arbitrary products of open sets?”. The topology that this is describing is called the box topology. Namely, the box topology has the set of all arbitrary products of open open sets as an open base. It is clear that the box and product topologies coincide for finite cases but do not for infinite cases. So, why did we choose the box topology over the product topology? The answer comes from the following theorem.

Theorem: Let $f:M\mapsto X$ be a mapping of a topological space $M$ into the product space $X$. Then, $f$ is continuous if and only if $\pi_jf:M\mapsto X_j$ is continuous for all $j$.

Proof:

$\implies$:

Let $G$ be open in $X_j$ then $\left(\pi_j f\right)^{-1}\left(G\right)=f^{-1}\left(\pi_j^{-1}(G)\right)$ and since $\pi_j^{-1}(G)$ is subbasic open in $X$ and $f$ continuous we see that $f^{-1}\left(\pi_j^{-1}(G)\right)$ is open in $M$.

$\Leftarrow$:

Lemma: A mapping $f:E\mapsto F$ (where $E$ and $F$ are arbitrary topological spaces) is continuous if and only if $f^{-1}(B)$ is open for every $B\in\mathcal{B}_F$ where $\mathcal{B}_F$ is an open base for $F$.

Proof:

$\implies$: Let $G$ be open in $F$ then $\displaystyle G=\bigcup_{h\in\mathcal{H}}B_h$ where $\left\{B_h\right\}_{h\in\mathcal{H}}\subseteq\mathcal{B}_F$. So then, $\displaystyle f^{-1}\left( G\right)=$$\displaystyle f^{-1}\left(\bigcup_{h\in\mathcal{H}}B_h\right)=\bigcup_{h\in\mathcal{H}}f^{-1}\left(B_h\right)$ which by assumption is the union of open sets in $E$ and thus open. The conclusion follows.

$\Leftarrow$: This is obvious. $\blacksquare$

So, suppose that $G$ is basic open in $X$, then $\displaystyle G=\pi_{j_1}^{-1}(G_{j_1})\cap\cdots\cap\pi_{j_m}^{-1}(G_{j_m})$ where $\left\{j_,\cdots,j_m\right\}\subseteq\mathcal{J}$ and $G_{j_k}$ is open in $X_k,\text{ }1\leqslant k\leqslant m$. It follows then that

$f^{-1}(G)$

$=f^{-1}\left(\pi_{j_1}^{-1}(G_{j_1})\cap\cdots\cap\pi_{j_m}^{-1}(G_{j_m})\right)$

$=$$f^{-1}\left(\pi_{j_1}^{-1}(G_{j_1})\right)\cap\cdots\cap f^{-1}\left(\pi_{j_m}^{-1}(G_{j_m})\right)$

$=\left(\pi_{j_1}f\right)^{-1}(G_{j_1})\cap\cdots\cap \left(\pi_{j_m}f\right)^{-1}(G_{j_m})$

and since each $\pi_jf,\text{ }j\in\mathcal{J}$ is continuous it follows that $f^{-1}(G)$ is the finite intersection of open sets in $M$ and thus open. The conclusion follows. $\blacksquare$

Note that the key point here was that $G$ was the finite intersection of inverse images of open sets. If we had the box topology we could have had that $G$ was the infinite intersection of inverse images of open sets we would have no guarantee that $f^{-1}(G)$ would be open since it would be the infinite intersection of open sets…which we all know need not be open.

Now, one may ask “Topologically does it matter which order we product the sets?”. The answer is “no…for finite cases”

Theorem: Let $X_1,\cdots,X_n$ be a finite collection of topological spaces then $X_1\times\cdots\times X_n\approx X_{\sigma(1)}\times\cdots\times X_{\sigma(n)}$ where $\sigma$ is an permutation of $\{1,\cdots,n\}$

Proof: It is readily verified that the canonical mapping $f:X_1\times\cdots\times X_n\mapsto X_{\sigma(1)}\times\cdots\times X_{\sigma(n)}$ given by $(x_1,\cdots,x_n)\mapsto (x_{\sigma(1)},\cdots,x_{\sigma(n)})$ is a homeomorphism. $\blacksquare$

February 20, 2010